题目：
Determine whether an integer is a palindrome. Do this without extra space.
Some hints: Could negative integers be palindromes? (ie, -1)
If you are thinking of converting the integer to string, note the restriction of using extra space.
You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
There is a more generic way of solving this problem.
题解：
最开始是用Reverse Integer的方法做的，通过了OJ，不过那个并没有考虑越界溢出问题。遇见这种处理Number和Integer的问题，首要考虑的就是会不会溢出越界。然后再考虑下负数，0。还有就是要心里熟悉\的结果和%的结果。感觉这种题就是考察你对数字敏感不敏感（反正我是很不敏感）
有缺陷的代码如下：
1 public boolean isPalindrome(
int x) {
2 if(x<0)
3 return false;
4 5 int count = 0;
6 int testx = x;
7 while(testx!=0){
8 int r = testx%10;
9 testx = (testx-r)/10;
10 count++;
11 }
12 13 int newx = x;
14 int result = 0;
15 while(newx!=0){
16 int r = newx%10;
17 int carry = 1;
18 int times = count;
19 while(times>1){
20 carry = carry*10;
21 times--;
22 }
23 result = result+r*carry;
24 newx= (newx-r)/10;
25 count--;
26 }
27 28 return result==x;
29 30 }
另外一种方法可以避免造成溢出，就是直接安装PalidromeString的方法，就直接判断第一个和最后一个，循环往复。这样就不会对数字进行修改，而只是判断而已。
代码如下：
1 public boolean isPalindrome(
int x) {
2 // negative numbers are not palindrome 3 if (x < 0)
4 return false;
5 6 // initialize how many zeros 7 int div = 1;
8 while (x / div >= 10) {
9 div *= 10;
10 }
11 12 while (x != 0) {
13 int left = x / div;
14 int right = x % 10;
15 16 if (left != right)
17 return false;
18 19 x = (x % div) / 10;
20 div /= 100;
21 }
22 23 return true;
24 }
Reference：
http://www.programcreek.com/2013/02/leetcode-palindrome-number-java/
